Question

Constants Periodic Table Correct A neutron colides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to move off at an anglo 0-45. The neutron's initial speed is 3.3x10 m/s Part B Determine the speeds of the two particles, v, and after the colision Express your answers using two significant figures. Enter your answers numerically separated by a comma. VE ΑΣΦ + + Submit Previous Anuwe Begres ARTH * Incorrect; Try Again; 3 attempts remaining Provide Feedback Next >

Answer 1

Helium nuclecs mie Given initial speed the neutron of V = 3:3x105mts initial speed of helium núcleos Vie omys (at rest) neutron mn 1.675 x 10 of of 48 m. - 4 mm MHE 6.7x10-27 kg. -27 kg mass mass after collision MHE Before collision quilesines Tone mn m Не, O2 = 45° o The CosO2 Vancos on Theory Verona Wsina, An How in by the law e conservation of momentum horizontal direction momentum befone collision = momentum after collision in horizontal direction in horizontal direction man to = m i coso, & mie were cosa eqo 2

of momentum and the law of conservation in verticle direction. 0 + 0 = mnhsino, & mie vite sino eq and the law of conservation of energy Total energy before Total energy after Collision Collision by so 12 to I m vi + I mne vil = since mte - 4mm so mwa mn 1412 + 4412 Не 2 Q not = + 4 un + 4 ut eg. = math Am How from eq. Cosot O + (4m) (4 m) Wrocoso, = Van Coso, + 4 Vice Cos Oy putting known values- 3.3x105 Vn Coso, + 4 W Cos 45 3.3x105 = cos ou + 2.828 Vhe the - eq. 6

and from eq. mm visino, + (4 m.) w sino = 0 Maino, +4 ute sin q = 0 putting the known values sino, + 4 vite gin 45 =0 » Unsino, + 2.828 he = 0 \T sino, 2.828 vie eq. 6 adding the squares of eq. 6 and eq. 6. Vai cos204 + sin?( = (0.82% Ve - 33x10572 + (Q.828 vite ut Cos?0,75 im20,7 = 7.997 V2 + (3.3 X10% ]18.66 We xos + 7.997 V2 He 2012 (1) 15.994 vota +(33X105)-18.66 vite xlos Putting this value of wh in eq. ③ - (3:3x105)2 = 15.994 ve ne +62.3 x105)?–18:66 vi get your ne = 16x18.66 vite = 19.994 ve He

Vio = 78.66x105 T9.994 m/ n لود 9.33x 104 m/s He putting this value The eqe 3 (3.3x105 72 2012 of in + 4 x(9.33 x 10472 87.62X.108xY n 412 10.89x1010 n 2012 10 10.89 x 10 = 3.48 XIDO اور د 7.41 X 1010 m/s V - 2.72x 105 m/s M putting the values of Vite and very in 3:3X105 = 2.72x105 cos. + 28.828 X9.33x 104 » 2.72x105 cos o 3.3x105 2.644 105 > Cosa 0.66 X105 2.72X105

Q Cosi to 0.243 -> Q, = 75.93 so Past A³ 0, = 75.93 = 76 20,= 95.99'2 angle of the neutron o is 76 Part B » 8.72x105 ms 9,33x 104 m/s He are Thus speeds of two particles 2.72 x 105 myg and g. 33x10 4 m/ $